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Re: Why is y a local variable here?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg63189] Re: Why is y a local variable here?
*From*: Peter Pein <petsie at dordos.net>
*Date*: Sat, 17 Dec 2005 03:46:21 -0500 (EST)
*References*: <200512130841.DAA08255@smc.vnet.net> <976C0928-EE8E-4FE2-824E-FE1F2C3636EF@omegaconsultinggroup.com> <dnv444$oas$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Steven T. Hatton schrieb:
> On Thursday 15 December 2005 11:27, Omega Consulting wrote:
>
>>I have a hard time understanding the purpose of your example. You say
>>that you want to define
>>
>>sd[y_] = SDP[n - 1, y]
>>
>>in your function. But you want the global value of y to be used. If
>>so, why have y as a parameter of sd. If the global value comes in,
>>then sd will always return the same thing (n and y are fixed values).
>>
>>If you want the function to use the global value, don't make it a
>>parameter of sd.
>
>
> I didn't want Global`y to be used, I just didn't understand why it was being
> hidden (renamed) when code using almost the same construct was seeing the
> Global`y.
>
>
>>The other part of your question is where does y$ come from, why isn't
>>it y (which would then evaluate into the global y value). This is a
>>subtle side-effect of how Mathematica scoping works. It is somewhat
>>complicated and usually only appears when you nest one scoping
>>construct inside another. In this case, the = for sd inside the :=
>>for SDP. The details are in section 2.7.4 of the Mathematica Book
>>(Advanced Topic: Variables in Pure Functions and Rules). That
>>describes the situation in a much better way than I could reproduce
>>here.
>
>
> That was my only question. Thank you. I will need to read that section
> carefully a few times. I believe the replacement is due to there being an
> unbound variable "n" within the sd[y_]=SDP[n-1,y];, and that variable is
> changed during evaluation. The discussion seems to focus on pure functions,
> but I suspect the same rules apply to Modules and other block structures as
> well. That seems to be what the rather cryptic comment about lambda
> expressions is about.
>
> Here is another nuance of the situation. The following is the correct
> implementation of the algorithm. If I change it by making y local to the
> Module, it does not us "x" in the result. Instead, the result contains the
> instance variable of the Module.
>
> SDP[0, x_] := x
> SDP[n_Integer?Positive, x_] := Module[{sd, srp = Sqrt[Prime[n]], py},
> sd[y_] = SDP[n - 1, y];
> Expand[sd[x + srp] sd[x - srp]]]
>
> SDP[5, x]
>
> Maeder makes in interesting observation at the bottom of page 221 in
> _Programming in Mathematica_.
>
> Steven
>
Hi Steven,
if you want to save evaluation time with the sd[]-construct, this will fail,
because sd[] will be redefined with evevery call of SDP[]. In that case you
should follow the standard procedure:
SDP[0,x_]=x; SDP[n_,x_}:=SDP[n,x]=...
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